Number of laps=n. It takes J 11n-1 minutes to do n laps; it takes Y 15n-2 minutes to do n laps.

In the time it takes Y to complete n laps, J has completed n+1 laps, so 11(n+1)-1=15n-2.

11n+11-1=15n-2; 12=4n so n=3. Therefore the time is 15*3-2=43 minutes (or 11*4-1=43 minutes).

The graph below shows that they meet up after 43 minutes, J having completed 4 laps while Y completes 3.

The more general solution is that J completes n+k laps while Y completes n laps:

11(n+k)-1=15n-2; 4n=11k+1; 4n=(8+3)k+1; n-2k=3k/4+1/4; m=(3k+1)/4 where m is an integer, and n=m+2k.

For m to be an integer the right-hand side must be divisible by 4, so k=4a+1 for integer a; so k=1, 5, 9, etc.

The lowest value for k is 1, so lowest value for m=1, and n=m+2k=1+2=3, making the shortest time 15*3-2=43 minutes.

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